Q. 10. - Calculus

Question

The graph of the equation $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ is a hyperbola for

any nonzero constants A and B.

(a) What is the eccentricity of the hyperbola?

(b) Explain why the eccentricity, e, of a hyperbola is always greater than 1.

(c) What is $\underset{A\to 0}{\lim }$? What happens to the shape of the hyperbola as A → 0?

(d) What is $\underset{A\to \infty }{\lim }$? What happens to the shape of the hyperbola as A→∞?

Step-by-Step Solution

Answer

Part (a) The answer is $e=\frac{\sqrt{A^2+B^2}}{A}$

Part b) The foci of a hyperbola are further away from the center and even the vertices, resulting in greater eccentricity.

Part c) The answer is$\underset{A\to B}{\lim }\frac{\sqrt{A^2+B^2}}{A}=0$is elongated and flattens.

Part d) The answer is$\underset{A\to \infty }{\lim }\frac{\sqrt{A^2+B^2}}{A}=1$is more like a parabola.

Part (a) Step 1: The objective is to find out the eccentricity of the hyperbola?

The given equation of hyperbola $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$

Where $A,B$ are non zero constants

Any conic section can be defined as the locus of points with constant distances to a point and a line. That ratio is known as eccentricity, and it is commonly represented by the symbol $e$

The eccentricity of a hyperbola is defined as

$e=\frac{\sqrt{A^2+B^2}}{A}$

Part (b) Step 1: The objective is to explain why the eccentricity, e, of a hyperbola is always greater than 1.

The eccentricity of a hyperbola is greater than $1$

Hence, The foci of a hyperbola are further away from the center and even the vertices, resulting in greater eccentricity.

Part (c) Step 1: The objective is to find out the value of lim A → 0 e and write the shape of the hyperbola as A → 0

The eccentricity is given by $e=\frac{\sqrt{A^2+B^2}}{A}$

Then,

$$\begin{gathered} \underset{A\to 0}{\lim }\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{0^2+B^2}}{0} \\ =\infty \end{gathered}$$

Therefore, the answer is $\underset{A\to B}{\lim }\frac{\sqrt{A^2+B^2}}{A}=0$

The hyperbola elongates and flattens.

Part (d) Step 1: The objective is to find out the value of lim A → ∞ e and write the shape of the hyperbola as A → ∞

The eccentricity is given by $e=\frac{\sqrt{A^2+B^2}}{A}.$

Then,

$$\begin{gathered} \underset{A\to 0}{\lim }e=\underset{A\to \infty }{\lim }\frac{\sqrt{A^2+B^2}}{A} \\ \underset{A\to \infty }{\lim }\frac{\sqrt{A^2+B^2}}{A}=\frac{\sqrt{{\infty }^2+B^2}}{\infty } \\ =\frac{\infty }{\infty } \\ =1 \end{gathered}$$

Hence the answer is $\underset{A\to \infty }{\lim }\frac{\sqrt{A^2+B^2}}{A}=1$

The hyperbola becomes more like a parabola.